National 5PhysicsSQARevision NotesDynamicsProjectile MotionProjectile Motion

Projectile Motion (SQA National 5 Physics): Revision Note

Exam code: X857 75

Written by: Katie M

Reviewed by: Leander Oates

Updated on 31 October 2025

Projectile motion

A projectile is an object which is launched and then moves freely under gravity
- The trajectory (path) of a projectile consists of a vertical component and a horizontal component
- These components are independent of each other and must be evaluated separately
In the horizontal direction, there are no unbalanced forces, which means
- the horizontal velocity is constant
- the horizontal acceleration is zero
In the vertical direction, the gravitational force (weight) acts on the object, which means
- the vertical velocity changes
- the vertical acceleration is constant at $9.8 m s^{- 2}$ downwards

Projectile launched horizontally

A cannonball launched horizontally moves with projectile motion. Horizontal distance travelled each second is constant; vertical distance travelled each second increases at a constant rate. Combined results in a curved trajectory. — When an object is launched horizontally, it has been given a forward motion through the air, but it is also pulled downward by the force of gravity. This results in the path of the projectile being curved.

Horizontal motion

The constant horizontal velocity of a projectile can be determined using:

$v_{h} = \frac{s}{t}$

Where:
- $v_{h}$ = horizontal velocity, measured in metres per second (m s^-1)
- $s$ = horizontal range, measured in metres (m)
- $t$ = time, measured in seconds (s)

Velocity-time graph for horizontal motion

Graph of horizontal velocity vs time showing constant horizontal velocity v subscript h. The area equals the horizontal range s = v subscript h times t. — **The horizontal range of a projectile can be found from the area under the graph of horizontal velocity against time**

The area under the $v_{h}$ - $t$ graph is equal to the horizontal displacement $s$ (also called the range)
This is equivalent to

$s = v_{h} t$

Vertical motion

The vertical velocity of a projectile can be determined using:

$v_{v} = u_{v} + a t$

Where:
- $u_{v}$ = initial vertical velocity, measured in metres per second (m s^-1)
- $v_{v}$ = final vertical velocity, measured in metres per second (m s^-1)
- $a$ = acceleration, measured in metres per second squared (m s^-2)
- $t$ = time, measured in seconds (s)
The acceleration due to gravity has a constant value equal to $g = 9.8 m s^{- 2}$ (on Earth)
For projectiles which are projected horizontally, the initial vertical velocity is zero, so $u_{v} = 0$
The expression then becomes

$v_{v} = g t$

Velocity-time graph for vertical motion

Graph showing vertical velocity vs time with constant positive acceleration. Initial velocity is zero, final velocity is v subscript v. Area under the graph equals vertical height h = half times v subscript v times t. — **The vertical height of a projectile can be found from the area under the graph of vertical velocity against time**

The area under the $v_{v}$ - $t$ graph is equal to the vertical height $h$
This is equivalent to

$h = \frac{1}{2} v_{v} t$

Where:
- $h$ = vertical height fallen, measured in metres (m)

Worked Example

A motorcycle stunt rider moving horizontally takes off from a point 1.25 m above the ground, landing 10 m away as shown.

The stunt rider and motorcycle reach the ground 0.50 s after taking off.

For the stunt rider and motorcycle, calculate

(i) the horizontal velocity as they take off

(ii) the vertical velocity as they reach the ground.

The effects of air resistance can be ignored.

Answer:

(i)

Step 1: List the known quantities in the horizontal direction

Horizontal range, $s$ = 10 m
Time of flight, $t$ = 0.50 s

Step 2: Write down the relationship between horizontal velocity, range, and time of flight

$v_{h} = \frac{s}{t}$

Step 3: Calculate the horizontal velocity at take-off

$v_{h} = \frac{10}{0.50} = 20 m s^{- 1}$

(ii)

Step 1: List the known quantities in the vertical direction

Initial vertical velocity, $u_{v}$ = 0
Acceleration due to gravity, $a = g$ = 9.8 m s^-2
Time of flight, $t$ = 0.50 s

Step 2: Write down the relationship between vertical acceleration, velocity, and time of flight

$v_{v} = u_{v} + a t$

Step 3: Calculate the vertical velocity at landing

The least precise input value is 2 s.f.
The answer is also to 2 s.f.

$v_{v} = 0 + (9.8) (0.50)$

$v_{v} = = 4.9 m s^{- 1}$

Examiner Tips and Tricks

The key thing you need to remember about projectile motion is that the horizontal motion and vertical motion are independent of each other. This just means you must solve each component separately. The only quantity that is the same in both motions is the time of flight.

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